Thread: Graphing

I need to graph the function s(x)=200x+1500/0.02x^2+5

I cant figure out how to graph this function...I'm used to functions without (/) in it!

find critical point f'(x)=0 so you can find the increasing and decreasing intervals that may help and find the concavity of the curve you should first find f''(x)=0 after finding the points (critical points,and the points where f''(x)=0 )but them in the coordinates and find the point where the curve intersect x-axes (y=0) and y-axis (x=0) .....

Originally Posted by lisa1984wilson

I need to graph the function s(x)=200x+1500/0.02x^2+5

I cant figure out how to graph this function...I'm used to functions without (/) in it!

The / is an operator used for division.
Akin to + for addition and - for subtraction
and * or x for multiplication.

Is you equation this?



If yes then a few of the x,y values

y.value x.value
75205 1
19155 2
8398.3 3
2755 10
4192.5 20

Generate the x,y values from you function and then plot the data.

The equation is set up like a big fraction not division

Originally Posted by lisa1984wilson

I need to graph the function s(x)=200x+1500/0.02x^2+5

I cant figure out how to graph this function...I'm used to functions without (/) in it!

next time, use the proper grouping symbols to make your expression clear ...

s(x) = (200x+1500)/(0.02x^2+5)

note that the numerator has degree less than the denominator ...

s(x) is asymptotic to y = 0, the x-axis.

s(x) has a single root at x = -7.5 and is negative-valued for x < -7.5 ; positive valued for x > -7.5

y-intercept is (0,300)

since the denominator is always positive, there are no vertical asymptotes or removable discontinuities.



at x = 10, s(x) has a relative maximum because s'(x) changes sign from (+) to (-)

at x = -25, s(x) has a relative minimum because s'(x) changes sign from (-) to (+)

s''(x) will tell you about inflection points and concavity.